Answer:
38.7 °C.
The final temperature (reached by both copper and water) is 38.7 °C.
So the answer is probably C.
The heat energy lost by the metal piece is equal to the heat gained by the water in the cup. From this concept using calorimetric equation, their final temperature is equal to 30°C.
What is calorimetric equation?Calorimetric equation states the relation between heat energy absorbed or released by a system with the mass, specific heat capacity and temperature difference of the matter as written below:
q = m c Δ
Given, mass of copper piece = 0.50 kg
mass of water = 0.50 kg
c for Cu = 0.386 KJ/Kg °C
c for water = 0.418 KJ/Kg °C
The heat energy lost by copper metal is equal to the heat energy gained by water. Let T be the final temperature.
then,
0.50 kg × 0.386 KJ/Kg °C × (115°C - T) = 0.50 kg × 0.418 KJ/Kg °C × (T - 22°C).
2.09 (115°C - T) = 0.193 (T - 22°C).
2.09 T - 45.98 + 0.193 = 22.195
T = 30°C
Therefore, the final temperature of water and copper piece is 30 °C.
Find more on calorimetry:
https://brainly.com/question/1407669
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On a sunny day, a rooftop solar panel delivers 60 W of power to the house at an emf of 17 V. How much current flows through the panel
Answer:
3.53 amps
Explanation:
Given data
Power= 60W
Voltage= 17V
The expression relating current, power, and voltage is
P= IV
substitute
60= I*17
I= 60/17
I= 3.53 amps
Hence the current that flows is 3.53 amps
The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.
Answer:
[tex]26852726.19\ \text{N}[/tex]
[tex]57.52^{\circ}[/tex]
Explanation:
r = Radius of circle = 7 m
w = Width of dam = 60 m
h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]
A = Area = [tex]rw[/tex]
V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]
Horizontal force is given by
[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]
Vertical force is given by
[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]
Resultant force is
[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]
The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].
The direction is given by
[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]
The line of action is [tex]57.52^{\circ}[/tex].